Help me understand and remember this Theorem

I am really struggling with a theorem in my book (Tao; Analysis II).I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:Let be an open subset of and let be a continous differentiable function and a point in with and .Then there exists an open subset of which contains and there exists an open subset of that contains and a function such that , and:= x_1,...x_{n-1})\in U\}" alt="\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1}))x_1,...x_{n-1})\in U\}" />In other words is a graph of a function on . Furthermore is differentiable in and we have: for all .

I am really struggling with a theorem in my book (Tao; Analysis II).I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:Let be an open subset of and let be a continous differentiable function and a point in with and .Then there exists an open subset of which contains and there exists an open subset of that contains and a function such that , and:= x_1,...x_{n-1})\in U\}" alt="\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1}))x_1,...x_{n-1})\in U\}" />In other words is a graph of a function on . Furthermore is differentiable in and we have: for all .

I'm not precisely sure what you want! It has a lot of uses. Intuitively it says that if you have an implicit equation in and the function satisfies the above conditions then you can "solve" the equation for in the sense that . Does that help?

I'm not precisely sure what you want! It has a lot of uses. Intuitively it says that if you have an implicit equation in and the function satisfies the above conditions then you can "solve" the equation for in the sense that . Does that help?

Hi Drexel28, first of all thanks for trying to help me. Any help is appreciated because I think that I don't really understand the theorem at all.

I get what you say. This is what I encounter when I google implicit functions. But doesn't the theorem say much more than that? And is the theorem really needed for expressing your z in a g(x,y) form?
[I am going to restate the theorem for n=3 and see if I can grasp it then].

The Implicit Function Theorem for :Let be an open subset of and let be a continous differentiable function and a point in with and .Then there exists an open subset of which contains and there exists an open subset of that contains and a function such that , and:= x_1,x_2)\in U\}" alt="\{(x_1,x_2,g(x_1,x_2))x_1,x_2)\in U\}" />In other words is a graph of a function on . Furthermore is differentiable in and we have: for all .

Hi Drexel28, first of all thanks for trying to help me. Any help is appreciated because I think that I don't really understand the theorem at all.

I get what you say. This is what I encounter when I google implicit functions. But doesn't the theorem say much more than that? And is the theorem really needed for expressing your z in a g(x,y) form?
[I am going to restate the theorem for n=3 and see if I can grasp it then].

No, you're one-hundred percent correct. It does say more than that, but that's the first thing one usually thinks of. Can you think of a specific part of the statement that you take issue with?

I am really struggling with a theorem in my book (Tao; Analysis II).I don't think I can properly remember this theorem because it contains so much, plus I dont think I really get the Theorem. So my question to you guys is, can you make this theorem understandable for me. Maybe give some simple examples. Show it's usefulness. Tell me how to remember it, etc.

Any help is really appreciated.

The Implicit Function Theorem:Let be an open subset of and let be a continous differentiable function and a point in with and .Then there exists an open subset of which contains and there exists an open subset of that contains and a function such that , and:= x_1,...x_{n-1})\in U\}" alt="\{(x_1,...,x_{n-1},g(x_1,...,x_{n-1}))x_1,...x_{n-1})\in U\}" />In other words is a graph of a function on . Furthermore is differentiable in and we have: for all .

Apologies in advance if this is too simple-minded to be useful. But the implicit function theorem is essentially a geometric result, and it's important to visualise the geometry before you worry about the analytic statement.

Start with the elementary, two-dimensional, case. The equation represents a curve in the plane. If at a point on the curve, then the tangent at that point is vertical and it may not be possible to express y uniquely as a function of x in a neighbourhood of that point. But if then y can be expressed as a function of x in some neighbourhood of , and in that neighbourhood.

Simplest possible example: . The curve is the unit circle, and the tangent is vertical at the points , where . In a neighbourhood of any other point on the circle, you can express y (locally) as a function of x, if , or if . What's more, the derivative is given by .

Now move on to three dimensions. The equation represents a surface in 3D-space. If at a point on the surface, then the vertical line at that point is tangent to the surface and it may not be possible to express z uniquely as a function of x and y in a neighbourhood of that point. But if then z can be expressed as a function of x and y in some neighbourhood of , and in that neighbourhood.

Exercise: work through the "simplest possible example", the unit sphere, given by the function . The setup is exactly analogous to the 2D example of the unit circle. if z lies on the "equator" of the sphere then and z cannot locally be expressed uniquely in terms of x and y. In a neighbourhood of any other point on the sphere, there is such an expression for z, and the partial derivatives and are given by and .

I think that if you can retain the picture of that simple example, then the implicit function theorem takes on a much more friendly appearance.

Opalg, thank you very much for your explanation. It really helps. The way you portrayed makes me understand the theorem better because I did not trip over open subsets in your posts. But the hardest part for me is to understand why this must be true: . What does it mean if (geometrically)?

I already figured that if it is non-zero then the Jacobian will be invertible (for the Inverse function theorem).